在实例上调用时找不到模板类中的模板成员函数 [英] Template member function in a template class not found when called on an instance

问题描述

namespace {
enum class API { CPU, CUDA };

template <API Api>
class Allocator {
    void* malloc(int size) const;
    void free(void* ptr) const;

    template <typename T>
    T* alloc1(int num) const {
        return static_cast<T*>(malloc(num * static_cast<int>(sizeof(T))));
    }
};

template <typename T, API Api>
T* alloc2(const Allocator<Api> allocator, int num) {
    return static_cast<T*>(allocator.malloc(num * static_cast<int>(sizeof(T))));
}

template <>
class Allocator<API::CPU> {
public:
    void* malloc(int size) const { return nullptr; }

    void free(void* ptr) const {}
};

Allocator<API::CPU> allocator;
int* ptr1 = allocator.template alloc1<int>(1);

int* ptr2 = alloc2<int>(allocator, 1);

}

alloc1 调用不会编译错误

.../src/test.cc:29:32: error: no member named 'alloc1' in '(anonymous namespace)::Allocator<API::CPU>'
int* ptr1 = allocator.template alloc1<int>(1);
                               ^
.../src/test.cc:29:38: error: expected unqualified-id
int* ptr1 = allocator.template alloc1<int>(1);
                                     ^
.../src/test.cc:29:42: error: expected '(' for function-style cast or type construction
int* ptr1 = allocator.template alloc1<int>(1);
                                      ~~~^

删除模板没有帮助:

.../src/test.cc:29:33: error: expected '(' for function-style cast or type construction
int* ptr1 = allocator.alloc1<int>(1);
                             ~~~^
.../src/test.cc:29:23: error: no member named 'alloc1' in '(anonymous namespace)::Allocator<API::CPU>'; did you mean 'malloc'?
int* ptr1 = allocator.alloc1<int>(1);
                      ^~~~~~
                      malloc

我正在使用符合C ++ 14标准的Clang 9.

I am using Clang 9 with C++14 standard.

alloc2 是一种解决方法，所以我只想知道为什么 alloc1 没有，或者什么是声明/调用它的正确方法.

alloc2 works as a workaround, so I just want to know why alloc1 doesn't or what is the proper way to declare/call it.

事实证明，提供了 Allocator< API :: CPU> :: malloc / free 的专业化，而不是 Allocator< API的专业化::CPU> 给出了我期望的行为，但在我的原始情况下，继承可能更合适.

it turns out that providing specializations of Allocator<API::CPU>::malloc/free instead of a specialization of Allocator<API::CPU> gives the behavior I expected, but probably inheritance will be more appropriate in my original situation.

推荐答案

class Allocator< API :: CPU> 专业化只是没有定义成员函数；它仅在通用模板中定义.如果要避免对多个指定重复相同的定义，则可以使用继承.例如:

class Allocator<API::CPU> specialisation simply does not define the member function; It is only defined in the generic template. If you want to avoid repeating same definition for multiple specilisations, you can use inheritance. Like this for example:

struct BaseAllocator {
    template <typename T>
    T* alloc1(int num) const {
        ...
    }
};

template <API Api>
class Allocator : public BaseAllocator {
    ...
};

template <>
class Allocator<API::CPU> : public BaseAllocator {
    ...
};

P.S.您的 alloc1 是私有的，因此无论如何您都无法从类外部访问它.

P.S. Your alloc1 is private, so you wouldn't have access to it from outside the class anyway.

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